Resolução – Física I – Mecânica – Sears; Zemansky

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Although rounded to three figures, this conversion is exact because the given conversion from inches to centimeters defines the inch.

?1000cm3 ? ? 1in ?3 1.2: 0.473 L ?ù ? ? ?ù ? ? = 28.9 in 3 . ? 1 L ? ? 2.54 cm ?

1.3: The time required for light to travel any distance in a vacuum is the distance divided by the speed of light;

3.00 ?ù108 m s g ? 1kg ? ?100cm?3 kg 1.4: 11.3 ?ù? ??ù? ? =1.13 ?ù 104 . cm3 ?1000 g ? ? 1 m ? m3

3 ? 1000 L ? ? 1 gal ? ? 128 oz. ? ? 1 bottle ? 1.6: 1 m3 ?ù ? 3 ? ?ù ? ? ?ù ? ??ù ? ?. ? 1 m ? ? 3.788 L ? ? 1 gal ? ? 16 oz. ? = 2111.9 bottles ? 2112 bottles The daily consumption must then be bottles ? 1 yr ? bottles yr ? 365.24 da ? da

furlongs ? 1 mile ? ? 1 fortnight ? ? 1 day ? mi 1.8: 180,000 ?ù ? ? ?ù ? ? ?ù ? ? = 67 . fortnight ? 8 furlongs ? ? 14 day ? ? 24 h ? h

? mi ? ? 1h ? ? 5280 ft ? ft 1.10: a) ? 60 ? ? ? ? ? = 88 ? hr ? ? 3600 s ? ? 1mi ? s ? ft ? ? 30.48 cm ? ? 1 m ? m b) ? 32 ? ? ? ? ? = 9.8 ? s 2 ? ? 1ft ? ? 100 cm ? s 2

? g??100cm?3?1kg? 3kg c) ?1.0 ? ? ? ? ? = 10 ? cm3?? 1m ? ?1000g? m3

1.11: The density is mass per unit volume, so the volume is mass divided by density. V=(60?ù103g)(19.5gcm3)=3077cm3 4 Use the formula for the volume of a sphere, V = ?r 3 , 3 to calculate r : r = (3V 4? ) = 9.0 cm 1/3

1.12: (3.16?ù107s-??ù107s)(3.16?ù107s)?ù100=0.58%

10 m – 890 ?ù103 m b) Since the distance was given as 890 km, the total distance should be 890,000 To report the total distance as 890,010 meters, the distance should be given as 890.01 km.

0.13%. b) If the chemical balance can measure to the nearest milligram, the error will be about 8.3?ù10-3%. c) If a handheld stopwatch (as opposed to electric timing devices) can measure to the nearest tenth of a second, the error will be about 2.8?ù10-2%.

The area is 9.69 ?? 0.07 cm2, where the extreme values in the piece’s length and 1.16: width are used to find the uncertainty in the area. The fractional uncertainty in the area is 0.07 cm2 = 0.72%, and the fractional uncertainties in the length and width are 9.69 cm 5.10 cm 1.9 cm

1.17: a) The average volume is ? (8.50 cm) (0.050 cm) = 2.8 cm3 2

4 (two significant figures) and the uncertainty in the volume, found from the extreme values of the diameter and thickness, is about 0.3 cm3 , and so the volume of a cookie is 2.8 ?? 0.3 cm3. (This method does not use the usual form for progation of errors, which is not addressed in the text. The fractional uncertainty in the thickness is so much greater than the fractional uncertainty in the diameter that the fractional uncertainty in the volume is 10% , reflected in the above answer.)

.05 1.18: (Number of cars ?ù miles/car.day)/mi/gal = gallons/day (2 ?ù 108 cars ?ù 10000 mi/yr/car ?ù 1 yr/365 days)/(20 mi/gal) = 2.75 ?ù 108 gal/day

1.19: Ten thousand; if it were to contain ten million, each sheet would be on the order of a millionth of an inch thick.

1.20: If it takes about four kernels to fill 1 cm3, a 2-L bottle will hold about 8000 kernels.

each page has between 500 and a thousand words (counting captions and the smaller print, such as the end-of-chapter exercise and problems), so an estimate for the number of words is about 106 .

1.22: Assuming about 10 breaths per minutes, 24 ?ù 60 minutes per day, 365 days per year, and a lifespan of fourscore (80) years, the total volume of air breathed in a lifetime isabout2?ù105m3.Thisisthevolumeofaroom100m?ù100m?ù20m,whichiskindof tight for a major-league baseball game, but it’s the same order of magnitude as the volume of the Astrodome.

1.24: With a pulse rate of a bit more than one beat per second, a heart will beat 105 times per day. With 365 days in a year and the above lifespan of 80 years, the number of beatsinalifetimeisabout3?ù109.With1L(50cm3)perbeat,andabout1gallonper 20 4 liter, this comes to about 4 ?ù107 gallons.

1.25: The shape of the pile is not given, but gold coins stacked in a pile might well be in the shape of a pyramid, say with a height of 2 m and a base 3 m ?ù 3 m . The volume of such a pile is 6 m3 , and the calculations of Example 1-4 indicate that the value of this volume is $6 ?ù108.

1.26: Thesurfaceareaoftheearthisabout4?R2=5?ù1014m2,whereRistheradiusof theearth,about6?ù106m,sothesurfaceareaofalltheoceansisabout4?ù1014m2.An averagedepthofabout10kmgivesavolumeof4?ù1018m3=4?ù1024cm3.Characterizing the size of a ÔÇ£dropÔÇØ is a personal matter, but 25 drops cm3 is reasonable, giving a total of 1026 drops of water in the oceans.

1.27: This will of course depend on the size of the school and who is considered a “student”. A school of thousand students, each of whom averages ten pizzas a year (perhaps an underestimate) will total 104 pizzas, as will a school of 250 students averaging 40 pizzas a year each.

bills can be stacked with about 2-3 per millimeter, so the number of bills in a stack to the moon would be about 1012. The value of these bills would be $1 trillion (1 terabuck).

(9,372,571km2?ù106m2km2)(15.6cm?ù6.7cm?ù1m2104cm2)=9?ù1014bills 1.30:

a) 11.1 m @ 77.6o b) 28.5 m @ 202o c) 11.1 m @ 258o d) 28.5 m @ 22o

1.35: r A = (12.0 m)sin 37.0o = 7.2 m, A = (12.0 m) cos 37.0o = 9.6 m. xy r = (15.0 m)cos 40.0o = = -(15.0 m) o = -9.6 m. B; B 11.5 m, B sin 40.0 xy

r (6.0 m) (6.0 m) xy A -1.00 m (a) tan ? = y = = -0.500 1.36: A 2.00 m X

?=tan-1(-0.500)=360o-26.6o=333o A 1.00 m (b) tan ? = y = = 0.500 A 2.00 m x

? = tan -1 (0.500) = 26.6o A 1.00 m (c) tan ? = y = = -0.500 A – 2.00 m x

? = tan (- 0.500) = 180o – 26.6o = 153o -1 A – 1.00 m (d) tan ? = y = = 0.500 A – 2.00 m x

second force has components F = F cos 32.4o = 433 N and F = F sin 32.4o = 275 N. 2x 2 2y 2 1x 1y F = F + F = 1158 N and F = F + F = 275 N x 1x 2 x y 1 y 2 y

r 1.39: Using components as a check for any graphical method, the components of B are r xy x a) The x – and y – components of the sum are 2.4 m and 10.8 m, for a magnitude ( )2 ( )2 ? 10.8 ? ? 2.4 ? c) The x- and y-components of the vector difference are ÔÇô 26.4 m and – 10.8 m, for a magnitude of 28.5 m and a direction arctan ( -10.8 ) = 202o. Note that -26.4 180o must be added to arctan( -10.8 ) = arctan(10.8 ) = 22o in order to give an angle in the – 26.4 26.4 third quadrant.

rr j (26.4 )2 (10 )2 ? 10.8 ? Magnitude = m + .8 m = 28.5 m at and angle of arctan? ? = 22.2o. ? 26.4 ?

1.40: Using Equations (1.8) and (1.9), the magnitude and direction of each of the given vectors is:

a) (-8.6 cm)2 + (5.20 cm)2 = 10.0 cm, arctan ( 5.20 ) = 148.8o (which is -8.60 180o ÔÇô 31.2o).

The total northward displacement is 3.25 km – 1.50 km = 1.75 km, , and the total westward displacement is 4.75 km . The magnitude of the net displacement is (1.75 km)2 + (4.75 km)2 = 5.06 km. The south and west displacements are the same, so The direction of the net displacement is 69.80o West of North.

1.42: a) The x- and y-components of the sum are 1.30 cm + 4.10 cm = 5.40 cm, 2.25 cm + (ÔÇô3.75 cm) = ÔÇô1.50 cm.

rr 1.43: a) The magnitude of A + B is ? ((2.80 ) + (1.90 ) )2 ? cm cos 60.0o cm cos 60.0o ? ? = 2.48 cm ? ((2.80 ) (1.90 ) )2 ? ? + cm sin 60.0o – cm sin 60.0o ? and the angle is ?(2.80cm)sin60.0o-(1.90cm)sin60.0o? arctan ? (2.80 ) (1.90 )cos 60.0o ? = 18 o

? cm cos 60.0o + cm ? rr b) The magnitude of A – B is

? ((2.80 cm) cos 60.0 – (1.90 cm) cos 60.0 )2 ? oo ? ? = 4.10 cm + ((2.80 cm) sin 60.0o + (1.90 cm) sin 60.0o )2 ?? ?? and the angle is ? (2.80 cm) sin 60.0o + (1.90 cm)sin 60.0o ? arctan ? (2.80 cm) cos 60.0o – (1.90 cm)cos 60.0o ? = 84 o

?? r r (r r) ? 1.44: A = (ÔÇô12.0 m) i^ . More precisely,

a) Ar = (3.60 m)cos 70.0o ^ + (3.60 m)sin 70.0o ^ = (1.23 m)i^ + (3.38 m) ^j 1.46: i j Br=-(2.40m)cos30.0oi^-(2.40m)sin30.0o^j=(-2.08m)i^+(-1.20m)^j b) = (3.00) r – (4.00) r r CAB =(3.00)(1.23m)i^+(3.00)(3.38m)^j-(4.00)(-2.08m)i^-(4.00)(-1.20m)^j =(12.01m)i^+(14.94)^j

(Note that in adding components, the fourth figure becomes significant.) c) From Equations (1.8) and (1.9), ( )2 ( )2 ? 14.94 m ? C = 12.01 m + 14.94 m = 19.17 m, arctan ? ? = 51.2o ? 12.01 m ?

= (4.00)2 + (3.00)2 = = (5.00)2 + (2.00)2 = 1.47: a) A 5.00, B 5.39

= 12 + 12 + 12 = 3 ? 1 so it is not a unit vector b) r A = A2 + A2 + A2 xyz

r If any component is greater than + 1 or less than ÔÇô1, A ? 1 , so it cannot be a unit r vector. A can have negative components since the minus sign goes away when the component is squared.

c) r A =1 2 (3.0)2 + 2 (4.0)2 = aa1 a 2 25 = 1 1 a = ?? = ??0.20 5.0

rr xy xy r r ( )i^ (A ) ^j A+B=A+B + +B xxyy

r r ( )i^ (B ) ^j B+A=B+A + +A xxyy rrrr Scalar addition is commutative, so A + B = B + A

rr A?B=AB+AB xx yy rr B?A=BA+BA xx yy rrrr Scalar multiplication is commutative, so A ? B = B ? A

b) r?ùr=(AB-AB)^+(AB-AB)^+(AB-AB)^ AB i j k yz zy zx xz xy yx

r ?ù r ( )^ ( ) ^ ( ) ^ BA=BA-BAi+BA-BAj+BA-BAk yz zy zx xz xy yx Comparison of each component in each vector product shows that one is the negative of the other.

ABcos?=(12m?ù15m)cos93o=-9.4m2 BCcos?=(15m?ù6m)cos80o=15.6m2 ACcos?=(12m?ù6m)cos187o=-71.5m2

Method 2: (Sum of products of components) A?B=(7.22)(11.49)+(9.58)(-9.64)=-9.4m2 B ? C = (11.49)(-3.0) + (-9.64)(-5.20) = 15.6 m2 A ? C = (7.22)(-3.0) + (9.58)(-5.20) = -71.5 m2

1.51: a) From Eq.(1.21), r r (4.00)(5.00) (3.00)(- ) A ? B = + 2.00 = 14.00.

1.52: For all of these pairs of vectors, the angle is found from combining Equations (1.18) and (1.21), to give the angle ? as rr ?A?B? ?AB+AB? ? AB ? ? AB ? In the intermediate calculations given here, the significant figures in the dot products and in the magnitudes of the vectors are suppressed.

rr a) A?B=-22,A=40,B=13,andso ? – 22 ? ? 40 13 ?

r r ? 60 ? ? 34 136 ? rr c) A ? B = 0, ? = 90.

1.54: a) From Eq. (1.22), the magnitude of the cross product is

(12.0 m)(18.0 m)sin (180o – 37o ) = 130 m2 The right-hand rule gives the direction as being into the page, or the ÔÇô z-direction. Using Eq. (1.27), the only non-vanishing component of the cross product is C=AB=(-12m)((18.0m)sin37o)=-130m2 z xy

b) The same method used in part (a) can be used, but the relation given in Eq. (1.23) gives the result directly: same magnitude (130 m2), but the opposite direction (+z-direction).

1.55: In Eq. (1.27), the only non-vanishing component of the cross product is

C = A B – A B = (4.00)(- 2.00) – (3.00)(5.00) = -23.00, z xy yx

rr 1.56: a) From the right-hand rule, the direction of A?ù B is into the page (the ÔÇô z-direction). The magnitude of the vector product is, from Eq. (1.22), Or, using Eq. (1.27) and noting that the only non-vanishing component is C=AB-AB z xy yx

= (2.80 cm)cos60.0o (-1.90 cm)sin 60o – (2.80 cm)sin 60.0o (1.90 cm)cos 60.0o = -4.61 cm2 gives the same result.

b) Rather than repeat the calculations, Eq. (1-23) may be used to see that rr ?ù has magnitude 4.61 cm2 and is in the +z-direction (out of the page). BA

1.57: a) The area of one acre is 1 mi ?ù 1 mi = 1 mi2 , so there are 640 acres to a square 8 80 640 mile.

? 1mi2 ? ?5280ft?2 b) (1 acre)?ù ? ? ?ù ? ? = 43,560 ft 2 ? 640 acre ? ? 1 mi ? (all of the above conversions are exact).

(43,560 ) ? 7.477 gal ? c) (1acre-foot)= ft3?ù? ?=3.26?ù105gal, ? 1 ft 3 ? which is rounded to three significant figures.

1.58:a) ($4,950,000102acres)?ù(1acre43560ft2)?ù(10.77ft2m2)=$12m2. c) $.008in2?ù(1in?ù78in)=$.007forpostagestampsizedparcel.

1.420 ?ù109 Hz ? cycles ? ? 3600 s ? cycles b) ?1.420 ?ù109 ? ?ù ? ? = 5.11?ù1012 ? s ? ? 1h ? h

c) Using the conversion from years to seconds given in Appendix F, ( ) ? 3.156 ?ù107 s ? ( ) ? 1y ?

1.60: Assume a 70-kg person, and the human body is mostly water. Use Appendix D to find the mass of one H O molecule: 18.015 u ?ù 1.661 ?ù 10ÔÇô27 kg/u = 2.992 ?ù 10ÔÇô26 2 kg/molecule. (70 kg/2.992 ?ù 10ÔÇô26 kg/molecule) = 2.34 ?ù 1027 molecules. (Assuming carbon to be the most common atom gives 3 ?ù 1027 molecules.

1.61: a) Estimate the volume as that of a sphere of diameter 10 cm: = 4 ? 3 = ?ù -4 3 V r 5.2 10 m 3 Mass is density times volume, and the density of water is 1000 kg m3 , so m=(0.98)(1000kg m3)(5.2?ù10-4m3)=0.5kg

b) Approximate as a sphere of radius r = 0.25?Ám (probably an over estimate) = 4 ? 3 = ?ù -20 3 V r 6.5 10 m 3 m=(0.98)(1000kg m3)(6.5?ù10-20m3)=6?ù10-17kg=6?ù10-14g

MM 1.62: a) ? = , so V = V? 0.200 kg x3= =2.54?ù10-5m3 7.86?ù103 kg/m3 x=2.94?ù10-2m=2.94cm

4?R3=2.54?ù105m3 – b) 3 R=1.82?ù10-2m=1.82cm

1.63: Assume each person sees the dentist twice a year for checkups, for 2 hours. Assume 2 more hours for restorative work. Assuming most dentists work less than 2000 hours per year, this gives 2000 hours 4 hours per patient = 500 patients per dentist. Assuming only half of the people who should go to a dentist do, there should be about 1 dentist per 1000 inhabitants. Note: A dental assistant in an office with more than one treatment room could increase the number of patients seen in a single dental office.

? 6.0 ?ù 1023 atoms ? ? 14 ?ù 10-3 kg ? mole b) The number of neutrons is the mass of the neutron star divided by the mass of a neutron: (2) (2.0?ù 1030 kg) (1.7 ?ù 10-27 kg neutron) c) The average mass of a particle is essentially 2 the mass of either the proton or 3

the neutron, 1.7 ?ù10-27 kg. The total number of particles is the total mass divided by this average, and the total mass is the volume times the average density. Denoting the density by ? (the notation introduced in Chapter 14).

4 ? R3? M (2?)(1.5?ù1011m)3(1018kgm3) = 3 = = 1.2 ?ù 1079. m 2 (1.7 ?ù 10-27 kg) ave m p 3

r r r r r r (r r r) A + B + C + D = 0, so D = – A + B + C A=+Acos30.0o=+86.6N,A=+Acos30.0o=+50.00N xy

B=-Bsin30.0o=-40.00N,B=+Bcos30.0o=+69.28N xy C=+Ccos53.0o=-24.07N,C=-Csin53.0o=-31.90N xy Then D = -22.53 N, D = -87.34 N xy

xy tan ? = D / D yx ? = 75.54o ? = 180o + ? = 256o , counterclockwise from + x – axis = 87.34 / 22.53

1.66: R = A + B = (170 km) sin 68o + (230 km) cos 48o = 311.5 km xxx R = A + B = (170 km) cos 68o – (230 km) sin 48o = -107.2 km yyy

R=R2+R2=(311.5km)2+(-107.2km)2=330km xy tan? = R R y 107.2 km = = 0.344 R x311.5 km

rrr r b) Algebraically, A = C – B, and so the components of A are A = C – B = (6.40 cm) cos 22.0o – (6.40 cm) cos 63.0o = 3.03 cm xxx yyy

xxxx =(12.0m)cos(90o-37o)+(15.00m)cos(-40o)+(6.0m)cos(180o+60o) = 15.7 m, and R=A+B+C yyyy

=(12.0m)sin(90o-37o)+(15.00m)sin(-40o)+(6.0m)sin(180o+60o) The magnitude of the resultant is R = R2 + R2 = 16.6 m , and the direction from xy

thepositivex-axisisarctan(-5.3)=-18.6o.Keepingextrasignificantfiguresinthe 15.7 intermediate calculations gives an angle of – 18.49??, which when considered as a positive counterclockwise angle from the positive x-axis and rounded to the nearest degree is 342o .

Take the east direction to be the x – direction and the north direction to be the y – direction. The x- and y-components of the resultant displacement of the first three displacements are then (-180m)+(210m)sin45o+(280m)sin30o=108m, -(210m)cos45o+(280m)cos30o=+94.0m,

keeping an extra significant figure. The magnitude and direction of this net displacement are ( )2 ( )2 ? 94 m ? ? 108 m ? The fourth displacement must then be 144 m in a direction 40.9o south of west.

The third leg must have taken the sailor east a distance (5.80 km) – (3.50 km) cos 45o – (2.00 km) = 1.33 km and a distance north (3.5 km)sin 45o = (2.47 km) The magnitude of the displacement is

(1.33 km)2 + (2.47 km)2 = 2.81 km andthedirectionisarctan(2.47)=62??northofeast,whichis90??-62??=28??east 1.33 of north. A more precise answer will require retaining extra significant figures in the intermediate calculations.

1.71: a) b) The net east displacement is -(2.80km)sin45o+(7.40km)cos30o-(3.30km)cos22o=1.37km,andthenetnorth displacementis-(2.80km)cos45o+(7.40km)sin30o-(3.30km)sin22.0o=0.48km, and so the distance traveled is (1.37 km)2 + (0.48 km)2 = 1.45 km.

(147 km)sin 85o + (106 km) sin 167o + (166 km)sin 235o = 34.3 km and the northward displacement is (147km)cos85o+(106km)cos167o+(166km)cos235o=-185.7km (A negative northward displacement is a southward displacement, as indicated in Fig. (1.33). Extra figures have been kept in the intermediate calculations.)

a) (34.3 km)2 + (185.7 km)2 = 189 km b) The direction from Lincoln to Manhattan, relative to the north, is

210 -10 second line is 42o + 30o = 72o. Therefore X = 10 + 250 cos 72o = 87 Y = 20 + 250 sin 72o = 258 b) The computer screen now looks something like this:

The length of the bottom line is (210 – 87)2 + (200 – 258) = 136 and its direction is 2

b) To use the method of components, let the east direction be the x-direction and the north direction be the y-direction. Then, the explorer’s net x- displacement is, in units of his step size, (40)cos 45o – (80)cos 60o = -11.7 and the y-displacement is (40)sin 45o + (80)sin 60o – 50 = 47.6.

The magnitude and direction of the displacement are ? 47.6 ? ? – 11.7 ? (More precision in the angle is not warranted, as the given measurements are to the nearest degree.) To return to the hut, the explorer must take 49 steps in a direction 104o – 90o = 14o east of south.

r 1.75: Let +x be east and +y be north. Let A be the displacement 285 km at 40.0o north r of west and let B be the unknown displacement.

rrr r A + B = R, where R = 115 km, east rrr B=R-A B =R -A B =R -A x x x, y y y

A=-Acos40.0o=-218.3km,A=+Asin40.0o=+183.2km xy R = 115 km, R = 0 xy xy

xy tan?=B B =(183.2km)(333.3km) yx ? = 28.8o , south of east

1.76: (a) ? = ? sin? par (b) ? = ? cos? perp (c) ? = ? sin? par

r r rrr E=R-B,E=R-B xxxyyy B=-Bsin43o=-158.2N,B=+Bcos43o=+169.7N xy R = 0, R = +132.5 N xy Then E = +158.2 N, E = -37.2N xy

direction, east the x-direction, and the z-axis vertical. The first displacement is then – 30k^, the second is – 15 ^j , the third is 200i^ (0.2 km = 200 m) , and the fourth is 100 ^j . Adding the four: – 30k^ – 15 ^j + 200i^ + 100 ^j = 200i^ + 85 ^j – 30k^

(b) The total distance traveled is the sum of the distances of the individual segments: 30 + 15 + 200 + 100 = 345 m. The magnitude of the total displacement is:

= 2 + 2 + 2 = 2 + 2 + (- )2 = D D D D 200 85 30 219 m xyz

r A = 240 m, 32o south of east rr B is 32o south of west and C is 62o south of west Let + x be east and + y be north rrr A+B+C=0 A + B + C = 0, so A cos 32o – B cos 48o – C cos 62o = 0 xxx A + B + C = 0, so – A sin 32o + B sin 48o – C sin 62o = 0 yyy A is known so we have two equations in the two unknowns B and C. Solving gives B = 255 m and C = 70 m.

1.80: Take your tent’s position as the origin. The displacement vector for Joe’s tent is (21cos23)^-(21sin23)^=19.33^-8.205^.ThedisplacementvectorforKarl’stentis oi oj i j (32cos37)^+(32sin37)^=25.56^+19.26^.Thedifferencebetweenthetwo oi oj i j displacements is: (19.33 – 25.56)i^ + (- 8.205 – 19.25) ^j = -6.23i^ – 27.46 ^j .

The magnitude of this vector is the distance between the two tents:

zz AB+AB=(Acos?)(Bcos?)+(Asin?)(Bsin?) xx yy A B A B = AB(cos ? cos ? + sin ? sin ? ) ABAB = AB cos( ? – ? ) AB = AB cos ? where the expression for the cosine of the difference between two angles has been used (see Appendix B).

r b) With A = B = 0, C = C ^ and C = C . From Eq. (1.27), k zz z z

C=AB-AB xx yx =(Acos?)(Bcos?)-(Asin?)(Bcos?) ABAA = AB cos ? sin ? – sin ? cos ? ABAB = AB sin(? – ? ) BA = AB sin ?

1.82: a) The angle between the vectors is 210o – 70o = 140o , and so Eq. (1.18) gives r r ( )( ) A?B=3.60m2.40mcos140o=-6.62m2Or,Eq.(1.21)gives rr A?B=AB+AB xx yy

=(3.60m)cos70o(2.4m)cos210o+(3.6m)sin70o(2.4m)sin210o = -6.62 m2

b) From Eq. (1.22), the magnitude of the cross product is (3.60m)(2.40m)sin140o=5.55m2,

and the direction, from the right-hand rule, is out of the page (the rr +z-direction). From Eq. (1-30), with the z-components of A and B vanishing, the z-component of the cross product is A B – A B = (3.60 m)cos 70o (2.40 m)sin 210o xy yx

Trianglearea=12(base)(height)=12(B)(Asin?) Parellogram area = BA sin ? b) 90o

1.84: With the +x-axis to the right, +y-axis toward the top of the page, and +z-axis out (r r) (r r) (r r) xyz

2 Ar – Br = (A – B )i^ + (A – B )^ + (A – B )k^ j xxyyzz b) = (- 5.00)i^ + (2.00) ^j + (7.00)k^

(5.00)2 + (2.00)2 + (7.00)2 = c) 8.83, rr and this will be the magnitude of B – A as well.

1.86: The direction vectors each have magnitude 3 , and their dot product is (1) (1) + (1) (ÔÇô1) + (1) (ÔÇô1) = ÔÇô1, so from Eq. (1-18) the angle between the bonds is arccos o

a restatement of the law of cosines. We know that C2=A2+B2+2ABcos?, rr where ? is the angle between A and B .

rr a)IfC2=A2+B2,cos?=0,andtheanglebetweenAandBis90o(thevectorsare perpendicular).

derive it. The most straightforward way, using vector algebra, is to assume the linearity of the dot product (a point used, but not explicitly mentioned in the text) to show that rr the square of the magnitude of the sum A + B is

( r r )? ( r r ) r r r r r r r r A+B A+B=A?A+A?B+B?A+B?B rrrrrr =A?A+B?B+2A?B rr = A2 + B2 + 2A ? B = A2 + B2 + 2AB cos?

Using components, if the vectors make angles ?A and ?B with the x-axis, the components of the vector sum are A cos ?A + B cos ?B and A sin ?A + B sin ?B, and the square of the magnitude is ( + )2 + (A sin ? + )2 Acos? Bcos? Bsin? A B A2(cos2A? B ) (cos2? + ) = +sin2? +B2 sin2? AABB + 2 AB(cos? cos ? + sin ? sin ? ) ABAB =A2+B2+2ABcos(?-?) AB =A2+B2+2ABcos? where ? = ?A ÔÇô ?B is the angle between the vectors.

rr rr b) A geometric consideration shows that the vectors A, B and the sum A + B rr must be the sides of an equilateral triangle. The angle between A, and B is 120o, since one vector must shift to add head-to-tail. Using the result of part (a), with A = B, the condition is that A2 = A2 + A2 + 2 A2cos ? , which solves 2

rr r to c and a to d as B,C, and D . In terms of unit vectors, rr()r() CjkDijk

Using Eq. (1.18), rr ? B ? D ? ? L2 ? arccos? ?=arccos?(L)(L )?=54.7o, ? BD ? ? 3 ? rr ? C ? D ? ? 2L2 ? ? ? ? ? ? 3?

1.90: From Eq. (1.27), the cross product is ? ? 6.00 ? 11.00 ? (-13.00) i^ + (6.00) ^j + (-11.00) k^ = 13 ?- (1.00 ) i^ + ? ? ^j – k^ ?. ? ? 13.00 ? 13.00 ?

The magnitude of the vector in square brackets is 1.93, and so a unit vector in this rr direction (which is necessarily perpendicular to both A and B) is

? – (1.00) i^ + (6.00 13.00) ^j – (11.00 13) ^ ? k ? 1.93 ?

The negative of this vector, ? (1.00) i^ – (6.00 13.00) ^j + (11.00 13) ^ ? k ? ?, ? 1.93 ?

r r rr 1.91: A and C are perpendicular, so A ? C = 0. A C + A C = 0, , which gives xx yy xy rr B ? C = 15.0, so – 3.5C + 7.0C = 15.0 xy We have two equations in two unknowns C and C . Solving gives xy xy

1.92: rr A?ùB = AB sin ? rr A ?ù B (- 5.00)2 + (2.00)2 sin ? = = = 0.5984 AB (3.00)(3.00) = sin-1(0.5984) = 36.8o ?

1.93: a) Using Equations (1.21) and (1.27), and recognizing that the vectors rr r A, B, and C do not have the same meanings as they do in those equations,

( r r ) r ( )i^ ( A B – A B ) ^ + (A B )k^ ) r A?ùB?C=AB-AB+ j -AB?C yz zy zx xz xy yx yzx zyx zxy xzy xyz yxz A similar calculation shows that r (r r) A?B?ùC=ABC-ABC+ABC-ABC+ABC-ABC xyz xzy yzx yxz zxy zyx and a comparison of the expressions shows that they are the same.

b) Although the above expression could be used, the form given allows for ready rr () compuation of A ?ù B the magnitude is AB sin ? = 20.00 sin 37.0o and the direction is, from the right-hand rule, in the +z-direction, and so ( r r ) r (20.00) (6.00) A ?ù B ? C = + sin 37.0o = +72.2.

(L + l) (W + w) = LW + lW + Lw, (L ÔÇô l) (W ÔÇô w) = LW ÔÇô lW ÔÇô Lw, where the common terms wl have been omitted. The area and its uncertainty are then WL ?? (lW + Lw), so the uncertainty in the area is a = lW + Lw.

b) The fractional uncertainty in the area is a lW + Wl l w = =+ A WL L W , the sum of the fractional uncertainties in the length and width.

c) The similar calculation to find the uncertainty v in the volume will involve neglecting the terms lwH, lWh and Lwh as well as lwh; the uncertainty in the volume is v = lWH + LwH + LWh, and the fractional uncertainty in the volume is v lWH + LwH + LWh l w h = =++, V LW H L W H

1.95: The receiver’s position is (+ 1.0 + 9.0 – 6.0 + 12.0)^ + (- 5.0 + 11.0 + 4.0 + 18.0) ^ = (16.0)^ + (28.0) ^. i jij

The vector from the quarterback to the receiver is the receiver’s position minus the quarterback’s position, or (16.0)^ + (35.0) ^ , a vector with magnitude ij (16.0)2 + (35.0)2 = 38.5, given as being in yards. The angle is arctan(16.0 ) = 24.6o to the 35.0 right of downfield.

ii) In AU, (1.3087)2 + (-.4423)2 + (-.0414)2 = 1.3820 iii) In AU,

c) The angle between the directions from the Earth to the Sun and to Mars is obtained from the dot product. Combining Equations (1-18) and (1.21), ? (-0.3182)(1.3087 – 0.3182) + (-0.9329)(-0.4423 – 0.9329) + ? = arccos ? ? (0.9857)(1.695)

d) Mars could not have been visible at midnight, because the Sun-Mars angle is less than 90o.

The law of cosines (see Problem 1.88) gives the distance as (138 )2 + (77 )2 + 2(138 )(77 ) o = ly ly ly ly cos154.4 76.2 ly,

where the supplement of 25.6o has been used for the angle between the direction b) Although the law of cosines could be used again, it’s far more convenient to use the law of sines (Appendix B), and the angle is given by ? sin 25.6o ? arcsin? 138ly?=51.5o,180o-51.5o=129o, ? 76.2 ly ? where the appropriate angle in the second quadrant is used.

r 1.98: Define S = Ai^ + B^j + Ck^ rr r?S=(xi^+y^+zk^)?(Ai^+B^j+Ck^) j = Ax + By + Cz rr r If the points satisfy Ax + By + Cz = 0, then r ? S = 0 and all points r are r perpendicular to S .

1.00?ù103m-63m,andsothemagnitudeoftheaveragevelocityis 1.00 ?ù103 m – 63 m 4.75 s 1.00?ù10 m = 169 m s 3 b) 5.90 s

2.2: a) The magnitude of the average velocity on the return flight is

(5150 ?ù 103 m) (13.5 da) (86, 400 s da) b) Because the bird ends up at the starting point, the average velocity for the round trip is 0.

2.3: Although the distance could be found, the intermediate calculation can be avoided by considering that the time will be inversely proportional to the speed, and the extra time will be ? 105 km hr ? ? 70 km hr ?

2.4: The eastward run takes (200 m 5.0 m s) = 40.0 s and the westward run takes (280 m 4.0 m s) = 70.0 s. a) (200 m + 280 m)/(40.0 s + 70.0 s) = 4.4 m s to two significant figures. b) The net displacement is 80 m west, so the average velocity is (80 m 110.0 s) = 0.73 m s in the ÔÇôx-direction (-i^).

2.5: In time t the fast runner has traveled 200 m farther than the slow runner: Slow runner has run (5.50 m s)t = 1570 m.

t = t + 33s sp d d = vt ? t = v dd = + 33s vv sp dd = + 33s 3.5 km 6.5 km ss d = 250 km

2.7: a) The van will travel 480 m for the first 60 s and 1200 m for the next 60 s, for a total distance of 1680 m in 120 s and an average speed of 14.0 m s . b) The first stage of the journey takes 240 m = 30 s and the second stage of the journey takes 8.0 m s (240 m 20 m s) = 12 s , so the time for the 480-m trip is 42 s, for an average speed of 11.4 m s. c) The first case (part (a)); the average speed will be the numerical average only if the time intervals are the same.

2.8: From the expression for x(t), x(0) = 0, x(2.00 s) = 5.60 m and x(4.00 s) = 20.8 m. a) 5.60m-0= 20.8m-0= 20.8m-5.60m= 2.80 m s b) 5.2 m s c) 7.6 m s 2.00 s 4.00 s 2.00 s

2.9: a) At t = 0, x = 0 , so Eq (2.2) gives 11 x (2.4 m s 2 )(10.0 s) 2 – (0.120 m s 3 )(10.0 s) 3 av t (10.0 s) 2

b) From Eq. (2.3), the instantaneous velocity as a function of time is

= – 2= 2 – 3 2 v 2bt 3ct (4.80 m s )t (0.360 m s )t , x

so i) v (0) = 0, x ii) v (5.0 s) = (4.80 m s )(5.0 s) – (0.360 m s )(5.0 s) 2 = 15.0 m s , 23 x

x c)Thecarisatrestwhenv=0.Therefore(4.80ms2)t-(0.360ms3)t2=0.The x

This is the time when the curve is most nearly straight and tilted upward (indicating postive velocity). c) V: Here the curve is plainly straight, tilted downward (negative velocity). d) II: The curve has a postive slope that is increasing. e) III: The curve is still tilted upward (positive slope and positive velocity), but becoming less so.

2.11: Time (s) 0 2 4 6 8 10 12 14 16 Acceleration (m/s2) 0 1 2 2 3 1.5 1.5 0

a) The acceleration is not constant, but is approximately constant between the times t = 4 s and t = 8 s.

2.12: The cruising speed of the car is 60 km hr = 16.7 m s . a) = 1.7 m s (to 16.7 m s 2 10 s two significant figures). b) 0-16.7 m s = -1.7 m s 2 c) No change in speed, so the acceleration 10 s is zero. d) The final speed is the same as the initial speed, so the average acceleration is zero.

((7.0 m s)t ) k^ r r (t ) = -(5.0 m s)ti^ + (10.0 m s)t^j + – (3.0 m s 2 )t 2

The velocity vector is the time derivative of the displacement vector:

r d r (t ) = (-5.0 m s)i^ + (10.0 m s) ^j + (7.0 m s – 2(3.0 m s 2 )t ) ^ k dt and the acceleration vector is the time derivative of the velocity vector:

r d 2 r (t ) = -6.0 m s2 k^ dt 2 At t = 5.0 s: ((7.0m s)(5.0s) ))k^ r r (t ) = -(5.0 m s)(5.0 s)i^ + (10.0 m s)(5.0 s) ^j + – (-3.0 m s 2 )(25.0 s 2 = (-25.0 m)i^ + (50.0 m) ^ – (40.0 m) ^ jk r d r (t ) = (-5.0 m s)i^ + (10.0 m s) ^j + ((7.0 m s – (6.0 m s 2 )(5.0 s)) k^ dt = (-5.0 m s)i^ + (10.0 m s) ^j – (23.0 m s) ^ k r d 2 r (t ) = -6.0 m s 2 k^ dt 2 (b) The velocity in both the x- and the y-directions is constant and nonzero; thus (c) The object’s acceleration is constant, since t does not appear in the acceleration vector.

dx 2.15:v==2.00cms-(0.125cms2)t x dt dv a = x = -0.125 cm s2 x dt xx x

c) Set x = 50.0 cm and solve for t. This gives t = 0 and t = 32.0s. The turtle returns d) Turtle is 10.0 cm from starting point when x = 60.0 cm or x = 40.0 cm. x

x Set x = 40.0 cm and solve for t : t = 36.4 s (other root to the quadratic equation is negative x

e) 2.16: Use of Eq. (2.5), with t = 10 s in all cases, a) ((5.0m/s)-(15.0m/s))/(10s)=-1.0m/s2 b) ((-15.0m/s)-(-5.0m/s))/(10s)=-1.0m/s2 c) ((-15.0m/s)-(-15.0m/s))/(10s)=-3.0m/s2.

2.17: a) Assuming the car comes to rest from 65 mph (29 m/s) in 4 seconds, x

b) Since the car is coming to a stop, the acceleration is in the direction opposite to the velocity. If the velocity is in the positive direction, the acceleration is negative; if the velocity is in the negative direction, the acceleration is positive.

(3.00 m s ) + (0.100 m s3 ) (0) = 3.00 m s , and the velocity at t = 5.00 s is (3.00 m s ) + (0.100 m s3 ) (5.00 s)2 = 5.50 m s , so Eq. (2.4) gives the average acceleration as (5.50 m s) – (3.00 m s) 2 (5.00 s)

b) The instantaneous acceleration is obtained by using Eq. (2.5),

dx v = = (9.60 m s )t – (0.600 m s )t 5 26 x dt dv 26 x dt There are two times at which v = 0 (three if negative times are considered), given by t = 0 and t4 = 16 s4. At t = 0, x = 2.17 m and ax = 9.60 m s 2. When t4 = 16 s4, x=(2.17m)+(4.80ms2) (16s4)ÔÇô(0.100)ms6)(16s4)3/2=14.97m, b)

2.21: a) Equating Equations (2.9) and (2.10) and solving for v0, 2(x – x ) 2(70 m) 0x x t 7.00 s b) The above result for v0x may be used to find v – v 15.0 m s – 5.00 m s a = x 0x = =1.43 m s , 2 x t 7.00 s or the intermediate calculation can be avoided by combining Eqs. (2.8) and (2.12) to eliminate v0x and solving for ax, ? v x – x ? ? 15.0 m s 70.0 m ? 2

0x v 2 ((173 mi hr) ( )) 2 0.4470 m s ax=2(x-x)= 2((307ft)(1m)) =32.0ms, x 1 mi hr 2 0 3.281ft

b) The time can be found from the above acceleration, v (173 mi hr) (0.4470 m s ) ax 32.0 m s2 The intermediate calculation may be avoided by using Eq. (2.14), again with v0x = 0, 2(x-x ) 2((307ft( 1m )) v (173 mi hr) 0.4470 m s x 1 mi hr

v = 0, a = v02 < a . Taking x = 0, FromEq.(2.13),with x x 2(x-x0) max 0 2.23: v2 ((105kmhr)(1ms)(3.6kmhr))2 2a 2(250 m s 2 ) max 2.24: In Eq. (2.14), with x ÔÇô x0 being the length of the runway, and v0x = 0 (the plane x t 8s 2.25: a) From Eq. (2.13), with v = 0, 0x v2 (20 m s)2 x 2(x - x ) 2(120 m) 0 0 0x x x-x=vt+1at2givesa=6.67fts2 0 0x 2 x x slowing down: 0 0x x x x 0x x 0 x 2 b) 1320 6.67 ft s , v 0 0x x x x 0x x 0 x x At 4.0 s, v = +2.7 cm s (to the right) At 7.0 s, v = -1.3 cm s (to the left) b)a=slopeofv-tgraph=-8.0cm/s=-1.3cms2whichisconstant 6.0 s c) x = area under v-t graph First 4.5 s: x=A +A Rectangle Triangle )?? ? 1 (4.5 s)?? 6 cm ?? = = (4.5 s 2 cm ? + 22.5 cm ?s?2 ?s? From 0 to 7.5 s: 1 ( )? cm ? 1 ( )? cm ? d= 6s?8 ?+ 1.5s?2 ?=25.5cm 2?s?2?s? 9s,theaccelerationisconstant(fromthegraph)andequalto45ms-20ms=6.3ms2.From 4s t = 9 s to t = 13 s the acceleration is constant and equal to 2 4s b) In the first five seconds, the area under the graph is the area of the rectangle, (20 m)(5 s) = 100 m. Between t = 5 s and t = 9 s, the area under the trapezoid is (1/2)(45 m/s + 20 m/s)(4 s) = 130 m (compare to Eq. (2.14)), and so the total distance in the first 9 s is 230 m. Between t = 9 s and t = 13 s, the area under the triangle is (1 2)(45 m s)(4 s) = 90 m , and so the total distance in the first 13 s is 320 m. 2.32: 2.33: a) The maximum speed will be that after the first 10 min (or 600 s), at which time the speed will be b) During the first 15 minutes (and also during the last 15 minutes), the ship will travel (1 2)(18 km s)(900 s) = 8100 km , so the distance traveled at non-constant speed is 16,200 km and the fraction of the distance traveled at constant speed is 16,200 km 1- = 0.958, 384,000 km c)Thetimespentatconstantspeedis384,000km-16,200km=2.04?ù104sandthetimespent 18 km s during both the period of acceleration and deceleration is 900 s, so the total time required forthetripis2.22?ù104s,about6.2hr. 1 (1.60ms2)(14.0s)2=156.8m 2 (from Eq. (2.12), with x0 = 0, v0x = 0), and has attained a speed of (1.60 m s2)(14.0s)=22.4 m s. During the 70-second period when the train moves with constant speed, the train travels (22.4 m s) (70 s) = 1568 m. The distance traveled during deceleration is given by Eq. (2.13), with v = 0,v = 22.4 m s and a = -3.50 m s2 , so the train moves a distance x 0x x x - x = -(22.4 m / s)2 = 71.68 m. The total distance covered in then 156.8 m + 1568 m + 71.7 m 0 2(-3.50 m/s2 ) = 1.8 km. In terms of the initial acceleration a1, the initial acceleration time t1, the time t2 during which the train moves at constant speed and the magnitude a2 of the final acceleration, the total distance xT is given by 1 1 (a t ) 2 ? a t ?? a t ? x = a t 2 + (a t )t + 1 1 = ? 1 1 ?? t + 2t + 1 1 ?, 2 2 | a2 | ? 2 ?? 2 T 11 11 2 1 | a2 | ? which yields the same result. c) d) From Fig. (2.35), the graphs have the same slope at t = 2 s . e) Car A passes car B when they have the same position and the slope of curve A is greater than that of curve B in Fig. (2.30); this is at t = 3 s. f) Car B passes car A when they have the same position and the slope of curve B is greater than that of curve A; this is at t = 1 s. truck's constant speed, and the car's position is given by xC = (1/2) aCt2. Equating the two expressions and dividing by a factor of t (this reflects the fact that the car and the truck are at the same place at t = 0) and solving for t yields 2v 2(20.0 m s) t = T = = 12.5 s aC 3.20 m s2 and at this time xT = xC = 250 m. b) aCt = (3.20 m/s2)(12.5 s) = 40.0 m/s (See Exercise 2.37 for a discussion of why the car's speed at this time is twice the truck's speed.) c) The car and the motorcycle have gone the same distance during the same time, so their average speeds are the same. The car's average speed is its constant speed vC, and for constant acceleration from rest, the motorcycle's speed is always twice its average, or 2vC. b) From the above, the motorcyle's speed will be vC after half the time needed to catch the car. For motion from rest with constant acceleration, the distance traveled is proportional to the square of the time, so for half the time one-fourth of the total distance has been covered, or d 4. 2.38: a) An initial height of 200 m gives a speed of 60 m s when rounded to one significant figure. This is approximately 200 km/hr or approximately 150 mi hr . (Different values of the approximate height will give different answers; the above may be interpreted as slightly better than order of magnitude answers.) b) Personal experience will vary, but speeds on the order of one or two meters per second are reasonable. c) Air resistance may certainly not be neglected. 2.39: a) From Eq. (2.13), with v = 0 and a = -g , yy v = 2g(y-y)= 2(9.80ms2)(0.440m)=2.94ms, 0y 0 which is probably too precise for the speed of a flea; rounding down, the speed is about b) The time the flea is rising is the above speed divided by g, and the total time is twice this; symbolically, 2g(y-y) 2(y-y) 2(0.440m) = 0= 0= = t 2 2 2 0.599 s, g g (9.80m/s2) or about 0.60 s. 2.40: Using Eq. (2.13), with downward velocities and accelerations being positive, v 2 = y (0.8 m s )2 + 2(1.6 m s 2 )(5.0 m) = 16.64 m 2 s 2 (keeping extra significant figures), so vy = 4.1 m s . d=(12)gt2,sot=2dg.Ifdismeasuredincentimeters,thereactiontimeis 22- g 980 cm s2 v+v2-2g(y-y) t = 0y 0y 0 g (5.00ms)+(5.00ms)-2(9.80ms2)(-12.0m) 2 = (9.80 m s 2 ) = 2.156 s, keeping an extra significant figure. The average velocity is then 12.0 m = 5.57 m s , down. 2.156 s As an alternative to using the quadratic formula, the speed of the ring when it hits the groundmaybeobtainedfromv2=v2-2g(y-y),andtheaveragevelocityfound y 0y 0 v +v from y 0 y ; this is algebraically identical to the result obtained by the quadratic formula. 2 b) While the ring is in free fall, the average acceleration is the constant acceleration due togravity,9.80m/s2down. 1 c) y = y + v t - gt 2 0 0y 2 1 0=12.0m+(5.00ms)t- (9.8ms2)t2 2 Solve this quadratic as in part a) to obtain t = 2.156 s. d)v2=v2-2g(y-y)=(5.00ms)2-2(9.8ms2)(-12.0m) y 0y 0 i) at t = 0.250 s, y = (40.0 m) + (5.00 m s )(0.250 s) ÔÇô (1/2)(9.80 m s 2 )(0.250 s)2 = 40.9 m, 2 vy = (5.00 m s ) ÔÇô (9.80 m s )(0.250 s) = 2.55 m s and ii) at t = 1.00 s, y = (40.0 m) + (5.00 m/s)(1.00 s) ÔÇô (1/2)(9.80 m/s2)(1.00 s)2 = 40.1 m, 2 vy = (5.00 m s ) ÔÇô (9.80 m s )(1.00 s) = ÔÇô 4.80 m s . b) Using the result derived in Example 2.8, the time is (5.00 m s) + (5.00 m s)2 - 2(9.80 m s2)(0 - 40.0 m) t = = 3.41 s. 2 (9.80 m s ) c) Either using the above time in Eq. (2.8) or avoiding the intermediate calculation by using Eq. (2.13), v2=v2-2g(y-y)=(5.00ms)2-2(9.80ms2)(-40.0m)=809m2s2, y 0y 0 vy = 28.4 m s . y 0y 11 b)y=vt-gt2=(-6.00ms)(2.00s)-(9.80ms2)(2.00s)2=-31.6m,withthe 0y 22 c) v2=v2-2g(y-y)=(6.00ms)2-2(9.80ms2)(-10.0m)=232m2s2,sov=15.2m y 0y 0 y 2.46: a) The vertical distance from the initial position is given by 1 0y 2 solving for v0y, y 1 (-50.0 m) 1 2 0y t 2 (5.00 s) 2 b) The above result could be used in v2 = v2 - 2g(y - y ), with v = 0, to solve for y y 0y 0 ÔÇô y0 = 10.7 m (this requires retention of two extra significant figures in the calculation 2 for v0y). c) 0 d) 9.8 m s , down. 2.47: a) b) c) The most direct way to 9.80 m s 2 ave d) (283 m s) (1.40 s) = but 40 g = 392 m s , so the figures are not consistent. 22 202 m s 2 2.48: a) From Eq. (2.8), solving for t gives (40.0 m s ÔÇô 20.0 m s )/9.80 m s = 2.04 s. b) Again from Eq. (2.8), 40.0 m s - (-20.0 m s) 2 9.80 m s c) The displacement will be zero when the ball has returned to its original vertical position, with velocity opposite to the original velocity. From Eq. (2.8), 40 m s - (-40 m s) 2 9.80 m s (This ignores the t = 0 solution.) 2 d) Again from Eq. (2.8), (40 m s )/(9.80 m s ) = 4.08 s. This is, of course, half the time found in part (c). the magnitude of g, and with g one-tenth as large, the height would be ten times higher, or 7.5 m. b) Similarly, if the ball is thrown with the same upward speed, it would go ten times as high, or 180 m. c) The maximum height is determined by the speed when hitting the ground; if this speed is to be the same, the maximum height would be ten times as large, or 20 m. v = v + ?t ?t dt 21 t1 ? = v + (t 2 - t 2 ) 11 2 ?? = v - t2 + t2 11 22 = (5.0 m s ) ÔÇô (0.6 m s3 )(1.0 s)2 + (0.6 m s3 ) t2 = (4.40 m s ) + (0.6 m s3 ) t2. At t2 = 2.0 s, the velocity is v2 = (4.40 m s ) + (0.6 m s3 )(2.0 s)2 = 6.80 m/s, or 6.8 m s b) From Eq. (2.16), the position x2 as a function of time is x = x + ? t v dt 2 1 t1 x =(6.0m)+?t((4.40m/s)+(0.6m/s3)t2)dt t1 3 (0.6 m s ) 11 3 At t = 2.0 s, and with t1 = 1.0 s, x = (6.0 m) + (4.40 m s )((2.0 s) ÔÇô (1.0 s)) + (0.20 m s3 )((2.0 s)3 ÔÇô (1.0 s)3) = 11.8 m. ?t A B v = ( At - Bt 2 ) dt = t 2 - t 2 = (0.75 m s 3 )t 3 - (0.040 m s 4 )t 3 x 0 23 ?t ? A B ? A B x = ? t 2 - t 3 ? dt = t 3 - t 4 = (0.25 m s 3 )t 3 - (0.010 m s 4 )t 4 . 0 ? 2 3 ? 6 12 b) For the velocity to be a maximum, the acceleration must be zero; this occurs at t=0 and t = A = 12.5 s . At t=0 the velocity is a minimum, and at t=12.5 s the velocity is B b) h = Area underv - tgraph max ?A +A Triangle Rectangle 1 ? cm ? ? (1.3 ms)?133 ? + (2.5 ms - 1.3 ms)(133 cm s) 2 ? s? ? 0.25 cm c) a = slope of vÔÇô t graph 133 cm s 2 a(0.5ms)?a(1.0ms)? =1.0?ù105cms 1.3 ms a (1.5 ms) = 0 because the slope is zero. d) h = area under vÔÇô t graph 1 ? cm ? -3 h(0.5ms)?A = (0.5ms)?33 ?=8.3?ù10 cm Triangle 2 ? s ? ? = 1 = ?ù -2 h (1.0 ms) A (1.0 ms)(100 cm s) 5.0 10 cm Triangle 2 vertical lines at t = 2.5s and t = 7.5s. This area is 1 (4.00 cm s2 + 8.00 cm s2 )(7.5 s - 2.5 s) = 30.0 cm s 2 2.54: a) To average 4 mi hr , the total time for the twenty-mile ride must be five hours, so the second ten miles must be covered in 3.75 hours, for an average of 2.7 mi hr . b) To average 12 mi hr , the second ten miles must be covered in 25 minutes and the average speed must be 24 mi hr . c) After the first hour, only ten of the twenty miles have been covered, and 16 mi hr is not possible as the average speed. The velocity and acceleration of the particle as functions of time are v(t)=(9.00ms3)t2-(20.00ms2)t+(9.00ms) x x b) The particle is at rest when the velocity is zero; setting v = 0 in the above expression and using the quadratic formula to solve for the time t, (20.0 m s3 ) ?? (20.0 m s3 )2 - 4(9.0 m s3 )(9.0 m s) t= 2(9.0 m s3 ) and the times are 0.63 s and 1.60 s. c) The acceleration is negative at the earlier time and positive at the later time. d) The velocity is instantaneously not changing when the acceleration is zero; solving the above expression for a (t) = 0 gives x 20.00 m s2 18.00 m s3 Note that this time is the numerical average of the times found in part (c). e) The greatest distance is the position of the particle when the velocity is zero and the acceleration is negative; this occurs at 0.63 s, and at that time the particle is at (3.00ms3)(0.63s)3-(10.0ms2)(0.63s)2+(9.00ms)(0.63s)=2.45m. (In this case, retaining extra significant figures in evaluating the roots of the quadratic equation does not change the answer in the third place.) f) The acceleration is negative at t = 0 and is increasing, so the particle is speeding up at the greatest rate at t = 2.00 s and slowing down at the greatest rate at t = 0. This is a situation where the extreme values of da a function (in the case the acceleration) occur not at times when = 0 but at the dt endpoints of the given range. 20.0 s 15 s c) Her net displacement is zero, so the average velocity has zero magnitude. d)50.0m=1.43ms.Notethattheanswertopart(d)istheharmonicmean,notthe 35.0 s arithmetic mean, of the answers to parts (a) and (b). (See Exercise 2.5). 2.57: Denote the times, speeds and lengths of the two parts of the trip as t1 and t2, v1 and v2, and l1 and l2. a) The average speed for the whole trip is l + l l + l (76 km) + (34 km) 1 2 = 1 2 = ( 76 km ) + ( 34 km ) = 82 km h, t + t (l v ) + (l v ) 1 2 1 1 2 2 88 km h 72 km h b) Assuming nearly straight-line motion (a common feature of Nebraska highways), the total distance traveled is l1ÔÇôl2 and v ave l - l (76 km) - (34 km) t +t 1 2 88 km h 72 km h ( 31.4 km hr to three significant figures.) 2.58: a) The space per vehicle is the speed divided by the frequency with which the cars 96 km h 2400 vehicles h An average vehicle is given to be 4.5 m long, so the average spacing is 40.0 m ÔÇô 4.6 m = 35.4 m. constant speed (5.1 s) as t2. The constant speed is then at1, where a is the unknown acceleration. The total l is then given in terms of a, t1 and t2 by 1 l = at 2 + at t , 1 12 2 and solving for a gives l (100 m) a = = = 3.5 m s2 . (12)t2+tt (12)(4.0s)2+(4.0s)(5.1s) 1 12 2.60: a) Simple subtraction and division gives average speeds during the 2-second b) The average speed increased by 1.6 m s during each 2-second interval, so the 2 acceleration is 0.8 m s . c) From Eq. (2.13), with v0 = 0, v = 2(0.8 m s )(14.4 m) = 4.8 m s. Or, 2 recognizing that for constant acceleration the average speed of 5.6 m/s is the speed one 2 second after passing the 14.4-m mark, 5.6 m s ÔÇô (0.8 m s )(1.0 s) = 4.8 m s . d) With both the acceleration and the speed at the 14.4-m known, either Eq. (2.8) e) From Eq. (2.12), x ÔÇô x0 = (4.8 m s )(1.0 s) + 1 (0.8 m s 2 )(1.0 s)2 = 5.2 m. This 2 is also the average velocity (1/2)(5.6 m s + 4.8 m s ) times the time interval of 1.0 s. 2.61: If the driver steps on the gas, the car will travel If the brake is applied, the car will travel (20ms)(3.0s)+(12)(-3.8ms2)(3.0s)2=42.9m, so the driver should apply the brake. m ?3651d??24h??3600s? 2.62: a) d=ct=(3.0?ù108 )(1y)? 4 ?? ?? ? s ? 1y ?? 1d ?? 1h ? = 9.5?ù1015 m b) d=ct=(3.0?ù108m)(10-9s)=0.30m s d 1.5?ù1011 m c) t = = = 500 s = 8.33 min c 3.0 ?ù108 m s d 2(3.84 ?ù108 m) d) t = = = 2.6 s c 3.0?ù108 m s d 3?ù109 mi e) t = = = 16,100 s = 4.5 h c 186,000 mi s be found from: v2=v2+2a(x-x)=0+2(1.6ms2)(30m)=96m2s2 A 0A A 0 v = 96 m2 s2 = 9.80 m s A A's time to cover the first 30 m is thus: v - v 9.80 m s t = A 0A = = 6.13 s aA 1.6 m s2 and A's total time for the race is: (350 - 30) m 6.13 s + = 38.8 s 9.80 m s B's speed at the end of 30 m is found from: v2=v2+2a(x-x)=0+2(2.0ms2)(30m)=120m2s2 B 0B B 0 v=120m2s2=10.95ms B B's time for the first 30 m is thus v - v 10.95 m s t = B 0B = = 5.48 s aB 2.0 m s2 and B's total time for the race is: (350 - 30) m 5.48s + = 34.7 s 10.95 m s 0x x 0x x x x This is the initial speed for the second 5.0 s of the motion. For the second 5.0 s: 0x x 0 x-x=vt+1at2gives150m=(25s2)a+(12.5s2)aanda=4.0ms2 0 0x 2 x x x x moves with constant velocity) is stationary. Then, the passenger train has an initial relative velocity of vrel,0 = 10 m s . This relative speed would be decreased to zero after the relative separation had decreased to vr2el,0 = + 500 m. Since this is larger in 2 arel magnitude than the original relative separation of 200 m, there will be a collision. b) The time at which the relative separation goes to zero (i.e., the collision time) is found by solving a quadratic (see Problems 2.35 & 2.36 or Example 2.8). The time is given by t = 1 (v - v2 + 2ax ) rel,0 rel,0 rel,0 a =(10s2m)(10ms-100m2s2-40m2s2) = (100 s)((1 - 0.6 ). Substitution of this time into Eq. (2.12), with x0 = 0, yields 538 m as the distance the passenger train moves before the collision. 2.67: The total distance you cover is 1.20 m + 0.90 m = 2.10 m and the time available is 1.20 m = 0.80 s . Solving Eq. (2.12) for ax, 1.50 m s ( x - x ) - v t (2.10 m) - (0.80 m s)(0.80 s) a = 2 0 0x = 2 = 4.56 m s . 2 20 m s 2.5 m s 2 At this time, the officer is v 2 (20 m s) 2 1 2a 2(2.5 m s2) This could also be found from (1/ 2)a1t12 , where t1 is the time found for the acceleration. At this time the car has moved (15 m s )(8.0 s) = 120 m, so the officer is 40 m behind the car. a) The remaining distance to be covered is 300 m ÔÇô x1 and the average speed is (1/2)(v1 + v2) = 17.5 m s , so the time needed to slow down is 360 m - 80 m = 16.0 s, 17.5 m s and the total time is 24.0 s. c) The officer slows from 20 m s to 15 m s in 16.0 s (the time found in part (a)), so 2 the acceleration is ÔÇô0.31 m s . TTT t = 2(40.0 m) = 6.17 s (2.10 m s) b) The car has moved a distance 1 a 3.40 m s2 a t 2 = C x = 40.0 m = 64.8 m, 2 C a 1 2.10 m s2 T 1 120 2 The cars collide when x1 = x2; setting the expressions equal yields a quadratic in t, 1 at 2 + v t - D = 0, 0 2 the solutions to which are t = v2 + 2aD - v , t = - v2 + 2aD - v 0000 aa The second of these times is negative and does not represent the physical situation. b) v = at = ( v 2 + 2aD - v ) 100 1 spreader when the brakes are applied, and the magnitude of the acceleration will be a=v12.Travellingat25ms,Juanisx=37m-(25ms)(0.80s)=17mfromthe 2 x1 2 spreader, and the speed of the car (and Juan) at the collision is obtained from ? v 2 ? ? x ? ? 17 m ? v 2 = v 2 - 2a x = v 2 - 2? 1 ? x = v 2 - v 2 ? 2 ? = (25 m s) 2 - (20 m s) 2 ? ? x 0x x 2 0x 2 0x 1 ? x1 ? ? ? ? ? 2 x 21 m 1 = 301 m2 s2 x b) The time is the reaction time plus the magnitude of the change in speed (v -v) 0 divided by the magnitude of the acceleration, or v - v 25 m s - 17.4 m s t = t + 2 0 x = (0.80 s) + 2 (21 m) = 1.60 s. flash reaction v 2 1 (20 m s) 2 0 2.72: a) There are many ways to find the result using extensive algebra, but the most straightforward way is to note that between the time the truck first passes the police car and the time the police car catches up to the truck, both the truck and the car have travelled the same distance in the same time, and hence have the same average velocity over that time. Since the truck had initial speed 3 v and the average speed is vp, the 2p 2p 3 3 b) The velocity is zero at t = ?? ? (a = 0 at t = 0, but this is an inflection point, not an ? extreme). The extreme values of x are then ? ? ? ?3 ? 2 ?3 ? ? 3 ?3? 3 ? ?? The positive value is then 1 2 ? (4.00 m s)3 ? 2 2 ? 3 3? 3 ? 2.00 m s ? 2.75: a) The particle's velocity and position as functions of time are v (t ) = v + ?0t ((-2.00 m s 2 ) + (3.00 3 )t ) dt ms x 0x ? 3? ? 3.00 m s ?t 2 , = 0 x (2.00 m s ? 2 ?? v- 2+ )t ? x(t ) = ?0t v (t )dt = v t - (1.00 m s 2 )t 2 + (0.50 m s 3 )t 3 x 0x = t (v - (1.00 m s 2 )t + (0.50 m s 3 )t 2 ), 0x where x0 has been set to 0. Then, x(0) = 0, and to have x(4 s) = 0, v - (1.00 m s 2 )(4.00 s) + (0.50 m s 3 )(4.00s) 2 = 0, 0x 0x x 2.76: The time needed for the egg to fall is 2 h 2(46.0 m - 1.80 m) t = = = 3.00 s, 9 (9.80 m s2) return. The total time is T = t + t = 2.5 s . Let y be the height of the building, then, 12 y = 1 gt 2 and y = v t There are three equations and three unknowns. Eliminate t , solve 2 1 s 2. 2 for t , and use the result to find y. A quadratic results: 1 gt 2 + v t - v T = 0. If 1 2 1 s1 s 2a Here, t = t , a = 1 2 g = 4.9 m s 2 , b = v = 340 m s , and c = -v T = -(340 m s)(2.5 s) = -850 m 1ss Then upon substituting these values into the quadratic formula, -(340ms)?? (340ms)2-4(4.9ms2)(-850m) t= 12 2(4.9 m s ) t1 = ( 2(4.9 m/s2 ) ) = 2.42 s - 340 m s )??(363.7 m s . The other solution, ÔÇô71.8 s has no real physical meaning. Then,y=1gt2=1(9.8ms2)(2.42s)2=28.6m.Check:(28.6m)(340ms)=.08s,the 212 time for the sound to return. 2.78: The elevators to the observation deck of the Sears Tower in Chicago move from the ground floor to the 103rd floor observation deck in about 70 s. Estimating a single (103)(3.5m) = floor to be about 3.5 m (11.5 ft), the average speed of the elevator is 5.15 m s. 70 s Estimating that the elevator must come to rest in the space of one floor, the acceleration 0 -(5.15 m s ) = - 2 22 2 3.5 m b) The required speed would be v=v2+2g(y-y)=(25ms)2+2(9.80ms2)(-21.3m)=14.4ms, 00 window is h ave 0 0 t The distance l from the roof to the top of the window is then v2 ((1.90 m) /(0.420 s) - (1 2)(9.80 m s2 )(0.420 s))2 l = 0 = = 0.310 m. 2g 2(9.80 m s2 ) An alternative but more complicated algebraic method is to note that t is the difference between the times taken to fall the heights l + h and h, so that 2(l + h) 2l gg Squaring the second expression allows cancelation of the l terms, (1 2)gt 2 + 2 gt 2l 2 = h, which is solved for 1 ? h ?2 l = ? - (1 2) gt ? , 2g ? t ? which is the same as the previous expression. 2 g = 2((9.80 m s)2 ) 2.81: a) The football will go an additional v 5.00 m s = 1.27 m above the window, so 2 2.83: a) From Eq. (2.14), with v0=0, yy0 b) The height above the release point is also found from Eq. (2.14), with v =7.59ms,v =0anda =-g, 0y y y v2 (7.59 m s)2 h = = = 2.94 m 0y 2g 2(9.80ms2) (Note that this is also (64.0 cm) ( ).The height above the ground is then 5.14 m. 2 45 m s g c) See Problems 2.46 & 2.48 or Example 2.8: The shot moves a total distance 2.20 m ÔÇô1.83 m = 0.37 m, and the time is 11 y= gt2= (9.8ms2)(9.0s2)=44.1m 22 To reach her ears after 3.0 s, the sound must therefore have traveled a total distance of h + (h - 44.1) m = 2h - 44.1 m ,where h is the height of the cliff. Given 340 m/s for the speed of sound: 2h - 44.1 m = (340 m s)(3.0 s) = 1020 m , which gives b)Wecanusev2=v2+2g(y-y)tofindtheteacher'sfinalvelocity.Thisgives y 0y 0 yy reaches an upward velocity of v=v+at=(5.0ms2)(10.0s)=50.0ms y 0y y andaheightofy=1at2=1(5.0ms2)(10.0s)2=250matthemomenttheengineisshut 2y 2 off. To find the helicopter's maximum height use v 2 = v 2 + 2a ( y - y ) y 0y y 0 Taking y = 250 m , where the engine shut off, and since v 2 = 0 at the maximum height: 0y -v2 y -y = 0y max 0 2g (50.0 m s)2 y = 250 m - = 378 m max 2(-9.8 m s2 ) or 380 m to the given precision. b) The time for the helicopter to crash from the height of 250 m where Powers stepped out and the engine shut off can be found from: 1 y=y+vt+at2=250m+(50.0ms)t+1(-9.8ms2)t2=0 0 0y 2 y 2 where we now take the ground as y = 0 . The quadratic formula gives solutions of t = 3.67 s and 13.88 s, of which the first is physically impossible in this situation. Powers' position 7.0 seconds after the engine shutoff is given by: 1 y=250m+(50.0ms)(7.0s)+ (-9.8ms2)(49.0s2)=359.9m 2 at which time his velocity is v = v + gt = 50.0 m s + (-9.80 m s )(7.0 s) = -18.6 m s 2 y 0y Powers thus has 13.88 - 7.0 = 6.88s more time to fall before the helicopter crashes, at his constant downward acceleration of 2.0m s2 . His position at crash time is thus: 1 y = y + v t + a t2 0 0y y 2 1 =359.9m+(-18.6ms)(6.88s)+ (-2.0ms2)(6.88s)2 2 = 184.6 m or 180 m to the given precision. Last 1.0 s of fall: y-y=vt+1at2givesh4=v(1.0s)+(4.9ms2)(1.0s)2 0 0y 2 y 0y Motion from roof to y - y = 3h 4 : 0 0y y v2=v2+2a(y-y)givesv=2(9.80ms2)(3h4)=3.834hms y 0y y 0 y This is vy for the last 1.0 s of fall. Using this in the equation for the first 1.0 s gives h 4 = 3.834 h + 4.9 Let h = u 2 and solve for u : u = 16.5. Then h = u 2 = 270 m. 2.88: a) t + t = 10.0 s fall sound return t + t = 10.0 s (1) fs d =d Rock Sound 1 gt 2 = v t f ss 2 1 (9.8 m s 2 )t 2 = (330 m s)t (2) fs 2 Combine(1)and(2):t=8.84s,t=1.16s fs m h = v t = (330 )(1.16 s) = 383 m ss s b) You would think that the rock fell for 10 s, not 8.84 s, so you would have thought it fell farther. Therefore your answer would be an overestimate of the cliff's height. 0 y 0y y - y = v t + 1 a t 2 gives v = 11.31 m s 0 0y 2 y 0y Use this v0y in v = v + a t to solve for v : v = -20.5 m s y 0y y y y b) Find the maximum height of the can, above the point where it falls from the scaffolding: 2 y 0y y 0 v 2 = v 2 + 2a ( y - y ) gives y - y = 6.53 m y 0y y 0 0 The can will pass the location of the other painter. Yes, he gets a chance. falling for a time t0 before Superman's leap (in this case, t0 = 5 s). Then, the height h of the building is related to t and t0 in two different ways: 1 - h = v t - gt 2 0y 2 = - 1 g (t + t )2 , 0 2 where v0y is Superman's initial velocity. Solving the second t gives t g - t0. = 2h hg Solving the first for v0y gives v0y = - + t, and substitution of numerical values gives t2 t = 1.06 s and v0y = ÔÇô165 m s , with the minus sign indicating a downward initial velocity. speed of the second part of the fall (with the Rocketeer supplying the upward acceleration), and assuming the student is a rest both at the top of the tower and at the v2 v2 ground, the distances fallen during the first and second parts of the fall are 1 and 1 , 2g 10g where v1 is the student's speed when the Rocketeer catches him. The distance fallen in free fall is then five times the distance from the ground when caught, and so the distance above the ground when caught is one-sixth of the height of the tower, or 92.2 m. b) The student falls a distance 5H 6 in time t = 5H 3g , and the Rocketeer falls the same distance in time tÔÇôt0, where t0=5.00 s (assigning three significant figures to t0 is more or less arbitrary). Then, 5H 1 = v (t - t ) + g (t - t ) 2 , or 00 0 62 5H 6 1 (t - t ) 2 00 0 At this point, there is no great advantage in expressing t in terms of H and g algebraically; 00 ? ? =?- 2 A takes the early lead. b) The cars are both at the origin at t = 0. The non-trivial solution is found by setting xA = xB, cancelling the common factor of t, and solving the quadratic for ] 2 2? Substitution of numerical values gives 2.27 s, 5.73 s. The use of the term ÔÇ£starting pointÔÇØ can be taken to mean that negative times are to be neglected. c) Setting vA = vB leads to a different quadratic, the positive solution to which is 2 6? d) Taking the second derivative of xA and xB and setting them equal, yields, 2? = 2? - 6?t . Solving, t = 2.67 s . position be x0. Note that these quantities are for separate objects, the student and the bus. The initial position of the student is taken to be zero, and the initial velocity of the bus is taken to be zero. The positions of the student x1 and the bus x2 as functions of time are then 10 20 a) Setting x = x and solving for the times t gives 12 1 (v ) t= ?? v2-2ax 000 a ((5.0 m s) )(40.0 m) ) 1 = ?? (5.0 m s)2 - 2(0.170 m s2 (0.170 m s2 ) = 9.55 s, 49.3 s. The student will be likely to hop on the bus the first time she passes it (see part (d) for a discussion of the later time). During this time, the student has run a distance 0 2= c) The results can be verified by noting that the x lines for the student and the bus intersect at two points: d) At the later time, the student has passed the bus, maintaining her constant speed, but the accelerating bus then catches up to her. At this later time the bus's velocity is (0.170 m s2 )(49.3 s) = 8.38 m s. quantities will be denoted symbolically. That is, 1 1( ) let y = h + v t - gt 2 , y = h - g t - t . In this case, t = 1.00 s . Setting 2 10200 22 y = y = 0, expanding the binomial (t - t )2 and eliminating the common term 12 0 2 0020 1 gt 2 t 1 gt - v 2 1 - 0 v 0 0 gt 0 Substitution of this into the expression for y and setting y = 0 and solving for h as a 11 function of v0 yields, after some algebra, ? 1 ?2 ? gt - v ? 0 2 0 (gt - v )2 00 a) Using the given value t = 1.00 s and g = 9.80 m s2 , 0 (4.9 )?? 4.9 m s - v ?2 ? 9.8 m s - v ? 0 This has two solutions, one of which is unphysical (the first ball is still going up when the second is released; see part (c)). The physical solution involves taking the negative 0 b) The above expression gives for i), 0.411 m and for ii) 1.15 km. c) As v0 approaches 9.8 m s , the height h becomes infinite, corresponding to a relative velocity at the time the second ball is thrown that approaches zero. If v > 9.8 m s , the first ball can never 0

catch the second ball. d) As v0 approaches 4.9 m/s, the height approaches zero. This corresponds to the first ball being closer and closer (on its way down) to the top of the roof when the second ball is released. If v > 4.9 m s , the first ball will already have 0

passed the roof on the way down before the second ball is released, and the second ball can never catch up.

equations h = 1 gt2, 2 h = 1 g(t – t)2 can be solved for t. Eliminating h and taking the 232

square root, t = 3 , and t = t , and substitution into h = 1 gt 2 gives h = 246 m. t – t 2 1- 2 3 2 This method avoids use of the quadratic formula; the quadratic formula is a generalization of the method of ÔÇ£completing the squareÔÇØ, and in the above form, 32

b) The above method assumed that t >0 when the square root was taken. The negative root (with t = 0) gives an answer of 2.51 m, clearly not a ÔÇ£cliffÔÇØ. This would correspond to an object that was initially near the bottom of this ÔÇ£cliffÔÇØ being thrown upward and taking 1.30 s to rise to the top and fall to the bottom. Although physically possible, the conditions of the problem preclude this answer.

(5.3m)-(1.1m) v = = 1.4 m s , x,ave (3.0 s) (-0.5 m) – (3.4 m) y ,ave (3.0 s)

b)v=(1.4ms)2+(-1.3ms)2=1.91ms,or1.9mstotwosignificantfigures, ave 1.4

3.2: a) x = (v ) t = (-3.8 m s)(12.0 s) = -45.6 m and x,ave r = y,ave b)r=x2+y2=(-45.6m)2+(58.8m)2=74.4m.

vij 3.5: a) (-170 m s) – (90 m s) b) a = = -8.7 m s , 2 x,ave (30.0 s) (40 m s) – (110 m s) 2 y ,ave (30.0 s)

-8.7 3.6: a) a = (0.45 m s ) cos 31.0?? = 0.39 m s , a = ) sin 31.0?? = 0.23 , 2222 (0.45 m s m s xy

b) r =?^-2?t^=(2.4ms)i^-[(2.4ms2)t]^j vij r 2 aj j r c) At t = 2.0 s , the velocity is = (2.4 m s) ^ – (4.8 m s) ^ ; the magnitude is vij (2.4ms)2+(-4.8ms)2=5.4ms,andthedirectionisarctan(-4.8)=-63??.The 2.4 acceleration is constant, with magnitude 2.4 m s 2 in the – y -direction. d) The velocity vector has a component parallel to the acceleration, so the bird is speeding up. The bird is turning toward the – y -direction, which would be to the bird’s right (taking the + z – direction to be vertical).

0y 0 b)vt=0.385mc)v=v=1.10ms,v=-gt=-3.43ms,v=3.60ms,72.2?? x x 0x y below the horizontal.

g b) The bomb’s constant horizontal velocity will be that of the plane, so the bomb x

c) The bomb’s horizontal component of velocity is 60 m/s, and its vertical component d)

e) Because the airplane and the bomb always have the same x-component of velocity and position, the plane will be 300 m above the bomb at impact.

v = 0, a = -9.80 m s , y – y = -h, t = 3.50 s 2 0y y 0 y – y = v t + 1 a t 2 gives h = 60.0 m 0 0y 2 y v = v sin 32.0??, y – y = -60.0 m, a = -9.80 m s = 2 0y 0 0 y y – y = v t + 1 a t 2 gives t = 3.55 s 0 0y 2 y 0x 0 x 0

3.12: Time to fall 9.00 m from rest: 1 y = gt 2 2 1 9.00m= (9.8ms2)t2 2 t=1.36s

Speed to travel 1.75 m horizontally: x=vt 0 1.75 m = v (1.36 s) 0

v = 1.3 m s 0 Use the vertical motion to find the time in the air: v = 0, a = -9.80 m s 2 , y – y = -(21.3 m – 1.8 m) = -19.5 m, t = ? 0y y 0 y – y = v t + 1 a t 2 gives t = 1.995 s 0 0y 2 y 0 x 0x 0 0x 2 x 0x b) v = 30.6 m s since a = 0 xx v = v + a t = -19.6 m s y 0y y

moment it leaves the tabletop and the time it will take for the ball to reach the floor (or rather, the rim of the cup). The latter can be determined simply by measuring the height of the tabletop above the rim of the cup and using y = 1 gt 2 to calculate the falling time. 2

The horizontal velocity can be determined (although with significant uncertainty) by timing the ball’s roll for a measured distance before it leaves the table, assuming that its speed doesn’t change much on the hard tabletop. The horizontal distance traveled while the ball is in flight will simply be horizontal velocity ?ù falling time. The cup should be placed at this distance (or a slightly shorter distance, to allow for the slowing of the ball on the tabletop and to make sure it clears the rim of the cup) from a point vertically below the edge of the table.

y 0y (15.0 m s) sin 45?? 2 9.80 m s b)UsingEquations(3.20)and(3.21)givesatt,(x,y)=(6.18m,4.52m): 1

23 c) Using Equations (3.22) and (3.23) gives at t , (v , v ) = (10.6 m s , 4.9 m s) : t , (10.6 m s , 0) t : (10.6 m s , – 4.9 m s), for 1xy 2 3 velocities, respectively, of 11.7 m s @ 24.8??, 10.6 m s @ 0?? and 11.7 m s @ -24.8??. Note that v is the same for all times, and that the y-component of velocity at t is x3 1

d) The parallel and perpendicular components of the acceleration are obtained from vrr vr v (a ? v )v v a ? v v v v || v 2 || v || Forprojectilemotion,v=-g^j, v?vr=-gv a so a , and the components of acceleration y

parallel and perpendicular to the velocity are t : -4.1 m s 2 , 8.9 m s 2 . t : 0, 9.8 m s 2 . 12 3

e) f) At t1, the projectile is moving upward but slowing down; at t2 the motion is instantaneously horizontal, but the vertical component of velocity is decreasing; at t3, the projectile is falling down and its speed is increasing. The horizontal component of velocity is constant.

0 b) Assuming a horizontal tabletop, v = 0 , and from Eq. (3.16), 0y 0x 0 c) On striking the floor, v = -gt = – 2gy = -3.83m s , and so the ball has a velocity y0 of magnitude 5.24 m s , directed 46.9?? below the horizontal.

d) Although not asked for in the problem, this y vs. x graph shows the trajectory of the tennis ball as viewed from the side.

3.17: The range of a projectile is given in Example 3.11, R = v2 sin 2? g . 00 a)(120ms)2sin110??(9.80ms2)=1.38km.b)(120ms)2sin110??(1.6ms2= ) 8.4 km .

g 9.80 m s2 2 2 2 y0 2g c) Regardless of how the algebra is done, the time will be twice that found in part (a), or 3.27 s d) v is constant at 20.0 m s , so (20.0 m s)(3.27 s) = 65.3 m . x

e) 3.19: a) v = (30.0 m s) sin 36.9?? = 18.0 m s ; solving Eq. (3.18) for t with y = 0 and 0y 0 y = 10.0 m gives (18.0ms)?? (18.0ms)2-2(9.80ms2)(10.0m) t = = 0.68 s, 2.99 s 2 9.80 m s b) The x-component of velocity will be (30.0 m s) cos 36.9?? = 24.0 m s at all times. The y-component, obtained from Eq. (3.17), is 11.3 m s at the earlier time and c) The magnitude is the same, 30.0 m s , but the direction is now 36.9?? below the horizontal.

b) The x-component of velocity is constant at v = (12.0 m s) cos 51.0?? = 7.55 m s . The x

y-component is v = (12.0 m s) sin 51.0?? = 9.32 m s at release and 0y

0y 0x f) 3.21: a) The time the quarter is in the air is the horizontal distance divided by the horizontal component of velocity. Using this time in Eq. (3.18), x gx 2 y-y =v – 0 0 y v 2v 2 0x 0x

gx 2 = tan? x – 0 2 2? v 2 cos 00

dart ? gd ? ? 2v2 cos2 ? dart 0 0 0?

Using the given values for d and ? to express this as a function of v , 00 ? 26.62 m 2 s 2 ? ? 2? ? v0 ? Then, a) y = 2.14 m , b) y = 1.45 m , c) y = -2.29 m . In the last case, the dart was fired with so slow a speed that it hit the ground before traveling the 3-meter horizontal 3.23: a) With v = 0 in Eq. (3.17), solving for t and substituting into Eq. (3.18) gives y

v v 2 sin ? (30.0 m/s) sin 33.0?? 2 2 22 ( y – y ) = 0 y = 0 0 = = 13.6 m 0 2g 2g 2(9.80 m/s2 )

b) Rather than solving a quadratic, the above height may be used to find the time the rock takes to fall from its greatest height to the ground, and hence the vertical component ofvelocity,v=2yg=2(28.6m)(9.80m/s2)=23.7m/s,andsothespeedofthe y

v cos?t = 45.0 m 0 45.0 m cos? = = 0.600 (25.0 m/s)(3.00 s)

? = 53.1?? b) v = (25.0 m/s) cos 53.1?? = 15.0 m/s x v =0 y v = 15.0 m/s a = 9.80 m/s2 downward c) Find y when t = 3.00 s y=vsin?t-1gt2 0 2 1 = (25.0 m/s)(sin53.1??)(3.00s) – (9.80 m/s 2 )(3.00 s) 2 2 = 15.9 m v = 15.0 m/s = constant x

v = v sin ? – gt = (25.0 m/s)(sin 53.1??) – (9.80 m/s 2 )(3.00 s) = -9.41 y0

0y y 0 y – y = v t + 1 a t 2 gives y – y = 296 m 0 0y 2 y 0

b) In 6.00 s the balloon travels downward a distance y – y = (20.0 s)(6.00 s) = 120 m . 0

c) The horizontal distance the rock travels in 6.00 s is 90.0 m. The vertical component of the distance between the rock and the basket is 176 m, so the rock is (176 m)2 + (90 m)2 = 198 m from the basket when it hits the ground.

d) (i) The basket has no horizontal velocity, so the rock has horizontal velocity 15.0 Just before the rock hits the ground, its vertical component of velocity is v = v + a t = 20.0 s + (9.80 m/s2 )(6.00 s) = 78.8 m/s , downward, relative to the ground. y 0y y The basket is moving downward at 20.0 m/s, so relative to the basket the rock has e) horizontal: 15.0 m/s; vertical: 78.8 m/s

3.26: a) horizontal motion: x – x = v t so t = 60.0 m 0 0 x (v0 cos 43??)t vertical motion (take + y to be upward): y-y=vt+1at2gives25.0m=(vsin43.0??)t+1(-9.80m/s2)t2 0 0y 2 y 0 2 Solving these two simultaneous equations for v and t gives v = 3.26 m/s and t = 2.51s . 00 b) v when shell reaches cliff: y

v = v + a t = (32.6 m/s) sin 43.0?? – (9.80 m/s 2 )(2.51 s) = -2.4 m/s y 0y y The shell is traveling downward when it reaches the cliff, so it lands right at the edge of the cliff.

Use the vertical motion to find the time it takes the suitcase to reach the ground: 0y 0 y 0 y – y = v t + 1 a t 2 gives t = 9.60 s 0 0y 2 y

square of the frequency, and hence inversely proportional to the square of the rotational period; tripling the centripetal acceleration involves decreasing the period by a factor of 3 , so that the new period T ? is given in terms of the previous period T by T ? = T / 3 .

3.29: Using the given values in Eq. (3.30), 4?2(6.38?ù106m) rad ((24 h)(3600 s/h)) 2 (Using the time for the siderial day instead of the solar day will give an answer that differs in the third place.) b) Solving Eq. (3.30) for the period T with a = g , rad

4?2(6.38?ù106m) 9.80 m/s2 3.30: 550 rev/min = 9.17 rev/s , corresponding to a period of 0.109 s. a) From Eq. (3.29), v = 2?R = 196 m/s . b) From either Eq. (3.30) or Eq. (3.31), T

rad 3.31: Solving Eq. (3.30) for T in terms of R and a , rad

3.32:a)UsingEq.(3.31),2?R=2.97?ù104m/s.b)EitherEq.(3.30)orEq.(3.31)gives T

rad 3.33: a) From Eq. (3.31), a = (7.00 m/s)2 /(15.0 m) = 3.50 m/s2 . The acceleration at the b) a = 3.50 m/s2 , the same as part (a), but is directed down, and still towards the center. c) From Eq. (3.29), T = 2?R / v = 2? (15.0 m)/(7.00 m/s) = 12.6 s .

rad tan b)

3.35: b) No. Only in a circle would a point to the center (See planetary motion in rad c) Where the car is farthest from the center of the ellipse.

3.36: Repeated use of Eq. (3.33) gives a) 5.0 = m/s to the right, b) 16.0 m/s to the left, and c) 13.0 = m/s to the left.

3.37: a) The speed relative to the ground is 1.5 m/s + 1.0 m/s = 2.5 m/s , and the time is 35.0 m/2.5 m/s = 14.0 s. b) The speed relative to the ground is 0.5 m/s, and the time is 70 s.

3.38: The walker moves a total distance of 3.0 km at a speed of 4.0 km/h, and takes a time of three fourths of an hour (45.0 min). The boat’s speed relative to the shore is 6.8 km/h downstream and 1.2 km/h upstream, so the total time the rower takes is 1.5 km 1.5 km 6.8 km/h 1.2 km/h 3.39: The velocity components are – 0.50 m/s + (0.40 m/s)/ 2 east and (0.40 m/s)/ 2 south, for a velocity relative to the earth of 0.36 m/s, 52.5?? south of west.

km/h, so the heading must be arcsin 80.8 = 14?? north of west. b) Using the angle found in 320 part(a),(320km/h)cos14??=310km/h.Equivalently, (320 km/h)2 – (80.0 km/h)2 = 310 km/h .

3.41: a) (2.0 m/s) 2 + (4.2 m/s) 2 = 4.7 m/s, arctan 2.0 = 25.5?? , south of east. 4.2 c)2.0m/s?ù190s=381m.

3.42: a) The speed relative to the water is still 4.2 m/s; the necessary heading of the boat is arcsin 2.0 = 28?? north of east. b) (4.2 m/s)2 – (2.0 m/s)2 = 3.7 m/s , east. d) 4.2 800 m/3.7 m/s = 217 s , rounded to three significant figures.

3.43: a) b) x : -(10 m/s) cos 45?? = -7.1 m/s. y := -(35 m/s) – (10 m/s)sin 45?? = -42.1 m/s . c) (-7.1 m/s) 2 + (-42.1 m/s) 2 = 42.7 m/s, arctan -42.1 = 80?? , south of west. -7.1

3.44: a) Using generalizations of Equations 2.17 and 2.18, v = v + ? t 3 , v = v + ?t – ? t 2 , and x = v t + ? t 4 , y = v t + t 2 – t 3 . b) Setting ?? x 0 x 3 y 0 y 2 0 x 12 0 y 2 6 v = 0 yields a quadratic in t, 0 = v + ?t – ? t 2 , which has as the positive solution y 0y 2 t = 1 [? + ? + 2v ? ]= 13.59 s, 2

?0 keeping an extra place in the intermediate calculation. Using this time in the expression for y(t) gives a maximum height of 341 m.

xy y b)Thespeedisv=(?2+4?2t2)1/2,dv/dt=0att=0.(Seepartdbelow.) c) r and v are perpendicular when their dot product is 0: (?t)(?)+(15.0m-?t2)?ù(-2?t)=?2t-(30.0m)?t+2?2t3=0.Solvethisfort: t=??(30.0m)(0.500m/s2)-(1.2m/s)2=+5.208s,and0s,atwhichtimesthestudentisat(6.25m, 22 2(0.500 m/s ) d) At t = 5.208 s , the student is 6.41 m from the origin, at an angle of 13?? from the x- axis.Aplotofd(t)=(x(t)2+y(t)2)1/2showstheminimumdistanceof6.41mat5.208 s:

e) In the x – y plane the student’s path is: r?? r 32

b) The positive time at which x = 0 is given by t 2 = 3? ? . At this time, the y-coordinate is ? 3?? 3(2.4 m/s)(4.0 m/s2 ) y = t 2 = = = 9.0 m 2 2? 2(1.6 m/s3) .

v2 ((88 km/h)(1m/s)/(3.6 km/h))2 a = = = 0.996 m/s2 ? 1 m/s2 2x 2(300 m) b) arctan( 15 m ) = 5.4?? . c) The vertical component of the velocity is 460 m-300 m (88km/h)(1m/s)15m=2.3m/s.d)Theaveragespeedforthefirst300mis44km/h,so 3.6 km/h 160 m the elapsed time is 300 m 160 m + = 31.1s, (44km/h)(1m/s)(3.6km/h) (88km/h)(1m/s)cos5.4??/(3.6km/h) or 31 s to two places.

a) The equations of motions are: 1 y = h + (v sin ?)t – gt 2 0 2 x = (v cos ?)t 0

v = v sin ? – gt y0 v = v cos ? x0 Note that the angle of 36.9o results in sin 36.9?? = 3/5 and cos 36.9?? = 4/5 . b) At the top of the trajectory, v = 0 . Solve this for t and use in the equation for y to y

= v sin ? = + (v sin ? ) – (v sin ? )2 find the maximum height: t 0 . Then, y h (v sin ?) 0 1 g 0 , which g 0g2g

=+22 = v sin ? ? = reduces to y h 0 . Using v 25gh / 8 , and sin 3 / 5 , this becomes 2g 0

y=h+(25gh/8)(3/5)2=h+9h,ory=25h.Note:Thisanswerassumesthaty=h.Taking 2 g 16 16 0 0 16 c) The total time of flight can be found from the y equation by setting y = 0 , assuming y = h , solving the quadratic for t and inserting the total flight time in the x equation to 0

find the range. The quadratic is 1 gt 2 – 3 v – h = 0 . Using the quadratic formula gives 2 50 ?? – 2 – 1 – ?? 9 ?À 25 gh + 16 gh = (3/5)v0 ( (3/5)v0) 4( g)( h) = = (3/5) 25gh/8 t 2 . Substituting v 25gh / 8 gives t 25 8 8 . 2( 1 g ) 0 g ( ) (3 ) 2

Collecting terms gives t: t = 1 9h ?? 25h = 1 h ?? 5 h . Only the positive root is 2 2g 2g 2 2g 2g

2g 0 8 5 2g

ground) and rearranging gives 2v 2 sin ? cos ? 2v 2 x2 – 0 0 0 x – 0x h = 0, gg The easier thing to do here is to recognize that this can be put in the form 2v0xv0y 2v2 x2 – x – 0x h = 0, gg the solution to which is v[ ] 0y 0y g x

3.53: The distance is the horizontal speed times the time of free fall, 2 y 2(90 m) x g (9.80 m/s2 )

0 0 x-y change in height. Substitution of numerical values gives v = 42.8 m/s . b) Using the 0

above algebraic expression for v in Eq. (3.27) gives 0 ? x ?2 y = x – ? ? (188.9 m) ? 188 m ? Using x = 116 m gives y = 44.1 m above the initial height, or 45.0 m above the ground, which is 42.0 m above the fence.

3.56:Theequationsofmotionsarey=(vsin?)t-1/2gt2andx=(vcos?)t,assuming 00 the match starts out at x = 0 and y = 0 . When the match goes in the wastebasket for the minimum velocity, y = 2D and x = 6D . When the match goes in the wastebasket for the maximum velocity, y = 2D and x = 7D . In both cases, sin ? = cos ? = 2 / 2. . To reach the minimum distance: 6D = 2 v t , and 2D = 2 v t – 1 gt 2 . Solving the first 20 202 equation for t gives t = 6D 2 . Substituting this into the second equation gives v0

2v 00 0

To reach the maximum distance: 7D = 2 v t , and 2D = 2 v t – 1 gt 2 . Solving the first 20 202 equation for t gives t = 7D 2 . Substituting this into the second equation v0

= – ( )2 gives2D7D1g7D2.Solvingthisforvgivesv=49gD/5=3.13gD,which, 2 v0 0 0

3.57: The range for a projectile that lands at the same height from which it was launched is R = v02 sin 2? , and the maximum height that it reaches is H = v02 sin ? . We must find R 2

g 2g

H= = sin?,D=6gDsin2? when D and v 6gD . Solving the height equation for , or 0 2g

a) Solving for v gives 0 gx2 / 2 cos2 ? x tan ? – y 0 0 0 b) Eliminating t between Equations 3.20 and 3.23 gives v as a function of x , y

v cos ? y00 00 Using the given values yields v = v cos ? = 8.28 m/s, v = -6.98 m/s, so x00 y

x, the solution to which is v2 cos ? ? 2gh ? x = 0 0 ?tan2? + ? g ? v02 cos ?0 ? 0 v cos ? [ ] 0000 g If h = 0 , the square root reduces to v sin ? , and x = R . b) The expression for x 00

becomesx=(10.2m)cos?+[sin2?+sin2?+0.98] 000 The angle ? = 90?? corresponds to the projectile being launched straight up, and there 0

is no horizontal motion. If ? = 0 , the projectile moves horizontally until it has fallen the 0

c) The maximum range occurs for an angle less than 45?? , and in this case the angle is about 36?? .

0 b) c) Using (14. m – 1.9 m) instead of h in the above calculation gives x = 6.3 m , so the man will not be hit.

3.61: a) The expression for the range, as derived in Example 3.10, involves the sine of twice the launch angle, and sin (2(90?? – ? )) = sin (180?? – 2? ) = sin 180?? cos 2? – cos 180?? sin 2? = sin 2? , 00000 and so the range is the same. As an alternative, using sin(90?? – ? ) = cos? and 0

cos(90??-?)=sin?intheexpressionfortherangethatinvolvestheproductofthesine 00 0

0 b) Again, the algebra is the same as that used in Problem 3.58; v = 8.4 m/s , at an angle x

A graph of y(t) vs. x(t) shows the trajectory of Mary Belle as viewed from the side:

d) In this situation it’s convenient to use Eq. (3.27), which becomes y=(1.327)x-(0.071115m-1)x2.Useofthequadraticformulagivesx=23.8m.

3.63: a) The algebra is the same as that for Problem 3.58, gx2 2cos2? (xtan? – y) 0 00 In this case, the value for y is – 15.0 m , the change in height. Substitution of numerical values gives 17.8 m/s. b) 28.4 m from the near bank (i.e., in the water!).

v2=v2cos2?+(vsin?-gt)2 0000 =v2(sin2? +cos2? )-2v sin? gt+(gt)2 00000

y 0y y 0 v2 = v2 + 2a ( y – y ) gives y – y = 81.6 m y 0y y 0 0 b) Both the cart and the rocket have the same constant horizontal velocity, so both travel the same horizontal distance while the rocket is in the air and the rocket lands in c) Use the vertical motion of the rocket to find the time it is in the air. 0y y y v = v + a t gives t = 8.164 s y 0y y 0 0x d) Relative to the ground the rocket has initial velocity components v = 30.0 m/s and 0x 0y e) (i)

Relative to the cart, the rocket travels straight up and then straight down (ii)

Time the ball is in the air: x (b) v (runner) = v (ball) xx 6.0 m/s = (20.0 m/s) cos? cos ? = 0.300 ? = 72.5??

y = v sin ?t – 1 gt 2 02 1 – 45.0 m = (20.0 m/s)(sin72.5??)t – (9.80 m/s 2 )t 2 2

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